In Fourier analysis, Parseval's Identity relates to "the summability of the Fourier series as a function."
In inner product space analysis, the "identity" works as a "Pythagorean theorem" relating the squared length of a vector to the sums of squares of its components.
How would the same theorem be central to two different applications? Or are they more or less connected because the "summability" of the Fourier Series, is somehow also based on a "Pythagorean" theorem?
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NO, summability in Pythagorean has nothing to do with with Parseval's theorem.
It just says that energy of the "system" is always same no matter how you look at it! So you calculate the energy in time domain, OR, you transform your function to another domain (e.g. frequency), the energy remains the same.
One way to look at it is that the preservation of basic physical law. For instance, if you want to calculate the kinetic energy of a system: $\frac{1}{2}m||v||^2$. Now you calculate the velocity in Cartesian coordinate/polar coordinate/cylindrical coordinate, the energy would remain the same, now matter how you look at the system. The same logic applies here!
No matter how one mess around a system, the basic physical laws always conserves itself. It's just a matter of perspective.
The Fourier theorem is in a way a special case of Parseval's identity. Consider the following: If you look at the space $L^2$ with inner product
$$\left\langle f,g \right\rangle = \frac{1}{2\pi} \int_{0}^{2\pi} f(x)\overline{g(x)} dx$$
Then the functions $e^{ikx}, k\in\mathbb{Z}$ form an orthonormal basis, so Parseval tells you $$\|f\|^2 = \sum_{k\in\mathbb{Z}} \left|\left\langle f,e^{ikx}\right\rangle\right|^2 $$
However $c_k = \left\langle f,e^{ikx}\right\rangle = \frac{1}{2\pi} \int_{0}^{2\pi} f(x)e^{-ikx} dx$ is exactly the $k$-th Fourier coefficient, so this is nothing other than Parseval's identity for Fourier series.
It all comes down to the fact, that taking the Fourier series of a function is nothing other then expressing this function in the basis given by $e^{ikx}$: $$ f = \sum_{k\in\mathbb{Z}} \left\langle f,e^{ikx}\right\rangle e^{ikx} = \sum_{k\in\mathbb{Z}} c_k e^{ikx} $$
(Your constants may vary depending on definitions)